PAT A1010 Radix (25分)

题目链接：https://pintia.cn/problem-sets/994805342720868352/problems/994805507225665536

题目描述
Given a pair of positive integers, for example, 6 and 110, can this equation 6 = 110 be true? The answer is yes, if 6 is a decimal number and 110 is a binary number.

Now for any pair of positive integers N​1​​ and N​2​​ , your task is to find the radix of one number while that of the other is given.

输入
Each input file contains one test case. Each case occupies a line which contains 4 positive integers:

N1 N2 tag radix

Here N1 and N2 each has no more than 10 digits. A digit is less than its radix and is chosen from the set { 0-9, a-z } where 0-9 represent the decimal numbers 0-9, and a-z represent the decimal numbers 10-35. The last number radix is the radix of N1 if tag is 1, or of N2 if tag is 2.

输出
For each test case, print in one line the radix of the other number so that the equation N1 = N2 is true. If the equation is impossible, print Impossible. If the solution is not unique, output the smallest possible radix.

样例输入
6 110 1 10

样例输出
2

代码

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long LL;
LL map[256];
LL inf = (1LL << 63) - 1;

void init() {
	for(char c = '0'; c <= '9'; c++)
		map[c] = c - '0';
	for(char c = 'a'; c <= 'z'; c++)
		map[c] = c - 'a' + 10;
}

LL convertNum10(char a[], LL radix, LL t) {
	LL ans = 0;
	int len = strlen(a);
	for(int i = 0; i < len; i++) {
		ans = ans * radix + map[a[i]];
		if(ans < 0 || ans > t)
			return -1;
	}
	return ans;
}

int cmp(char N2[], LL radix, LL t) {
	int len = strlen(N2);
	LL num = convertNum10(N2, radix, t);
	if(num < 0)
		return 1;
	if(t > num)
		return -1;
	else if(t == num)
		return 0;
	else 
		return 1;
}

LL binarySearch(char N2[], LL left, LL right, LL t) {
	LL mid;
	while(left <= right) {
		mid = (left + right) / 2;
		int flag = cmp(N2, mid, t);
		if(flag == 0) 
			return mid;
		else if(flag == -1)
			left = mid + 1;
		else
			right = mid - 1;
	}
	return -1;
}

int findLargestDigit(char N2[]) {
	int ans = -1, len = strlen(N2);
	for(int i = 0; i < len; i++) {
		if(map[N2[i]] > ans)
			ans = map[N2[i]];
	}
	return ans + 1;
}

char N1[20], N2[20], temp[20];
int tag, radix;
int main() {
	init();
	scanf("%s %s %d %d", N1, N2, &tag, &radix);
	if(tag == 2) {
		strcpy(temp, N1);
		strcpy(N1, N2);
		strcpy(N2, temp);
	}
	LL t = convertNum10(N1, radix, inf);
	LL low = findLargestDigit(N2);
	LL high = max(low, t) + 1;
	LL ans = binarySearch(N2, low, high, t);
	if(ans == -1)
		printf("Impossible\n");
	else
		printf("%lld\n", ans);
	return 0;
}